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ikadub [295]
3 years ago
15

Comment Both propane and benzene are hydrocarbons. As a rule,

Chemistry
1 answer:
kozerog [31]3 years ago
5 0

The enthalpy change : -196.2 kJ/mol

<h3>Further explanation  </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction

2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)

∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

\tt \Delta H_{rxn}=2.(-285.8)-2.(-187.8)\\\\\Delta H_{rxn}=-571.6+375.4=-196.2~kJ/mol\rightarrow \Delta Hf~O_2=0

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How many moles of MgSiO3 are in 237g of the compound?
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Why do substances with weak intermolecular forces form solids at low melting points?
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faust18 [17]

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Final temperature = 152.57K,

Pressure = 0.6907 bar.

dT/dt = - 1,151 K/s.

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NB; fT = final temperature, fP = final pressure and iT = initial temperature.

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Therefore, [fT]³⁰/₈.₃₁₄ = 109.52 × 10⁶.

Final temperature=  [fP]³⁰/₈.₃₁₄ × 169.05.

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R = 30 - 8.314 = 21.86 J/mol K.

Then, the rate of change of the gas temperature at this time = dT/dt = - 1,151 K/s.

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